Please allow me to recycle the notations and rephrase the problem in a way which is, I hope, more intuitive. Let $\Omega \subset \mathbb{R}^n$ be a bounded open set. For $r > 0$, set
$$ \Omega_r = \{ x \in \Omega \, : \, d(x, \partial \Omega) < r \} . $$
Then the claim is that:
$$ (*) \; \mbox{ For $0 < r_2 < r_1 < \infty$, } \; \mathrm{Vol}(\Omega_{r_1}) \le (r_1/r_2)^n \mathrm{Vol}(\Omega_{r_2}) . $$
Without loss of generality (dilating/contracting $\mathbb{R}^n$ if needed), we may assume $r_2 = 1$ and $r := r_1 > 1$.
I don't know how to prove $(*)$ simply 'by the coarea formula': some functions are to be chosen as inputs to the formula, then one needs some sufficient knowledge about (the integral of one function against) the measures on the level sets (of the other function).
A 'direct' (i.e., not explicitly relying on the coarea formula) and rigorous proof can be produced by refining the argument given in George Lowther's answer to this MSE post. However, for the sake of connecting this line of argument to the coarea formula, I shall sketch here a somewhat more heuristic argument 'proving' the 'claim' (**) below (which could be made absolutely rigourous under some regularity assumption on $\partial \Omega$).
'Proof of $(*)$':
Step 1 - Reduction to (**). Consider $u : \Omega \to [0, \infty), u(x) = d(x, \partial \Omega)$ (and w.l.o.g. we have $r \in \mathrm{Im}(u)$). It is not hard to prove that $u$ is a $1$-Lipschitz function whose gradient $\nabla u$ is almost everywhere defined and with norm $1$. (The a.e. existence of $\nabla u$ follows from Rademacher's theorem, or by direct analysis along the lines of Federer's Curvature Measures (section 4).) Applying the coarea formula to this $u$ and to $g \equiv 1$ over the domain $\Omega_r$ to get the second and last equalities below, we get
\begin{align}
\mathrm{Vol}(\Omega_r) &= \int_{\Omega_r} \| \nabla u(x)\| dx = \int_0^r \left(\int_{[u=s]} dH_{n-1}(x)\right)ds \\
&= \int_0^r \mathrm{Vol}_{n-1}([u=s])ds = r \int_0^1 \mathrm{Vol}_{n-1}([u=rv])dv \\
&\overset{(**)}{\le} r^n \int_0^1 \mathrm{Vol}_{n-1}([u=v])dv = r^n \mathrm{Vol}(\Omega_1).
\end{align}
Here, inequality $(**)$ follows from the 'claim':
$$ (**) \; \mbox{ For $1 < r < \infty$ and $0 < v < 1$, } \; \mathrm{Vol}_{n-1}([u=rv]) \le r^{n-1} \mathrm{Vol}_{n-1}([u=v]) . $$
Step 2 - 'Proof' of (**). Recall that the $d$-dimensional Hausdorff measure of $S \subset \mathbb{R}^n$ is
$$ \mathrm{Vol}_d(S) = \lim_{\delta \to 0^+} H^d_{\delta}(S) \, ; \; H^d_{\delta}(S) := \mathrm{inf} \left\{ \sum_{i=1}^{\infty} (\mathrm{diam} \, U_i)^d \, : \, \bigcup_{i=1}^{\infty} U_i \supseteq S, \; \mathrm{diam} \, U_i < \delta \right\}.$$
The intuitive idea is the following: for sufficiently small $\delta$, consider a cover $\{U_i\}_i$ of $[u=v]$ such that all $\mathrm{diam} \, U_i < \delta$ and $\sum_i (\mathrm{diam} \, U_i)^{n-1}$ approximates $H^{n-1}_{\delta}([u=v])$. For each $U_i$, take some point $x_i \in U_i \cap [u=v]$ (if no such $x$ exists, discard $U_i$ from the cover) and consider $y_i \in \partial \Omega$ such that $d(x_i,y_i) = v$. Consider the dilation $\phi_i : \mathbb{R}^n \to \mathbb{R}^n$ with center $y_i$ and ratio $r$, and set $U'_i = \phi_i(U_i)$. Of course, $\mathrm{diam} \, U'_i = r \, \mathrm{diam} \, U_i$.
If the choices are appropriate, the collection $\{U'_i\}_i$ should cover $[u=rv]$, in which case
\begin{align}
H^{n-1}_{r\delta}([u=rv]) \le \sum_i (\mathrm{diam} \, U'_i)^{n-1} = r^{n-1} \sum_i (\mathrm{diam} \, U_i)^{n-1} \approx r^{n-1} H^{n-1}_{\delta}([u=v]).
\end{align}
Taking $\delta \to 0$ would yield the 'claim' (**). 'QED'
Of course, for this argument to work, one needs a lot of fine tuning. In comparison, an approach more in the spirit of George Lowther's answer (where one would approximate $\Omega_1$ and $\Omega_r$ from within by (homothetic copies of) spherical sectors with apexes in $\partial \Omega$) requires coarser tuning.
